Disk View Factor from a Plate Element

Introduction

In this article, we will derive analytical view factor expressions for a disk from a plate element. We describe the disk geometry and relative position with respect to the plate element by parameters $(R, h, \omega)$ as shown in Figure 1. The actual view factor values can be calculated by our online tool: View Factor Calculator.

Figure 1: Geometrical Configuration of a Disk and a Plate Element for View Factor Evaluation.

View Factor Evaluation based on the Area Integration

A disk view factor from a plate element can be derived, by executing the area integration based on the view factor definition.

$$\begin{align} F &= \int_A \frac{\cos \Omega \cos \Lambda}{\pi S^2} dA = \int_0^R \int_0^{2\pi} \frac{\cos \Omega \cos \Lambda}{\pi (r^2+h^2)}\ r d \beta dr \notag \\ &= \int_0^R \int_0^{2\pi} \frac{r^2h \sin \omega \cos \beta + rh^2 \cos \omega}{\pi (r^2+h^2)^2} d\beta dr = \int_0^R \frac{2 rh^2 \cos \omega}{(r^2+h^2)^2} dr \notag \\ &= \left[- \frac{h^2 \cos \omega}{r^2 + h^2} \right]^R_0 = \frac{R^2}{R^2 + h^2} \cos \omega = \frac{1}{1 + (\frac{h}{R})^2} \cos \omega \end{align}$$

The parameter transformations for $\cos\Omega$ and $\cos\Lambda$ are performed as shown in Eqs. (2) and (3).

$$\begin{align} &\cos \Omega = \frac{\boldsymbol{\omega} \cdot \boldsymbol{r}}{\|\boldsymbol{\omega}\| \|\boldsymbol{r}\|} = \frac{r \sin \omega \cos \beta + h \cos \omega}{\sqrt{r^2 + h^2}} \\ &\mathrm{where}\quad\boldsymbol{\omega} = \left( \begin{array}{c} \sin\omega \\ 0 \\ \cos \omega \end{array} \right), \hspace{10pt} \boldsymbol{r} = \left( \begin{array}{c} r \cos \alpha \\ r \sin \alpha \\ h \end{array} \right) \notag \end{align}$$ $$\begin{equation} h \tan \Lambda = r, \quad \cos \Lambda = \frac{h}{\sqrt{r^2 + h^2}} \end{equation}$$

The calculation up to this point is the case where the entire disk is visible from the plate element. If the orientation of the plate element becomes more inclined (i.e. $\omega > \arctan \frac{h}{R}$), a part of the disk goes out of the view from the plate element. In this case, the area integration must be performed only for the visible part of the disk, which significantly complicates the calculation.

Figure 2: Disk View Factor Calculation by Area Integration.

To execute the area integration correctly, we divide the disk into two parts: $\textcircled{1}$ sector of the disk ($0 \le \beta \le \beta_0$), and $\textcircled{2}$ triangular area.

$$\begin{align} &F = \int_A \frac{\cos \Omega \cos \Lambda}{\pi S^2} dA \notag \\ &= \int_0^R \int_{0}^{\beta_0} \frac{2 \cos \Omega \cos \Lambda}{\pi (r^2+h^2)} r d \beta dr + \int^0_{R \cos \beta_0} \int_0^{x \tan \beta_0} \frac{2 \cos \Omega \cos \Lambda}{\pi (x^2 + y^2 + h^2)} dy dx \notag \\ &= \int^R_0 \int_{0}^{\beta_0} \frac{2r^2h \sin \omega \cos \beta + 2rh^2 \cos \omega}{\pi (r^2 + h^2)^2} d\beta dr + \int^0_{R \cos \beta_0} \int_0^{x \tan \beta_0} \frac{2xh \sin \omega + 2h^2 \cos \omega}{\pi (x^2 + y^2 + h^2)^2} dy dx \notag \\ &= \int^R_0 \frac{2r^2h \sin \omega \sin \beta_0}{\pi (r^2 + h^2)^2}dr + \int^R_0 \frac{2\beta_0 rh^2 \cos \omega}{\pi (r^2 + h^2)^2}dr + \int_0^{R \sin \beta_0} \int^{\frac{y}{\tan \beta_0}}_{R \cos \beta_0} \frac{2xh \sin \omega}{\pi (x^2 + y^2 + h^2)^2} dx dy \notag \\ &\hspace{11pt}+ \int^0_{R \cos \beta_0} \int_0^{x \tan \beta_0} \frac{2h^2 \cos \omega}{\pi (x^2 + y^2 + h^2)^2} dy dx \end{align}$$

For the parameter transformations of $\cos \Omega$ and $\cos \Lambda$ in the triangular area, we applied Eqs. (4) and (5).

$$\begin{align} &\cos \Omega = \frac{\boldsymbol{\omega} \cdot \boldsymbol{r}}{\|\boldsymbol{\omega}\| \|\boldsymbol{r}\|} = \frac{x \sin \omega + h \cos \omega}{\sqrt{x^2 + y^2 + h^2}} \\ &\mathrm{where}\quad \boldsymbol{\omega} = \left( \begin{array}{c} \sin\omega \\ 0 \\ \cos \omega \end{array} \right), \hspace{5pt} \boldsymbol{r} = \left( \begin{array}{c} x \\ y \\ h \end{array} \right) \notag \end{align}$$ $$\begin{gather} h \tan \Lambda = \sqrt{x^2 + y^2}, \quad \cos \Lambda = \frac{h}{\sqrt{x^2 + y^2 + h^2}} \end{gather}$$

Now, we evaluate each integral term in Eq. (4).

First term:

$$\begin{align} &\int^R_0 \frac{2r^2h \sin \omega \sin \beta_0}{\pi (r^2 + h^2)^2}dr = \frac{2 \sin \omega \sin \beta_0}{\pi} \int_0^{\frac{R}{h}} \frac{u^2}{(u^2 + 1)^2} du, \hspace{10pt} \mathrm{where}\ u=\frac{r}{h} \notag \\ &= \frac{2 \sin \omega \sin \beta_0}{\pi} \int_0^{\arctan \frac{R}{h}} \frac{\tan^2 t}{(\tan^2 t + 1)^2} \frac{dt}{\cos^2 t}, \hspace{10pt} \mathrm{where}\ \tan t=u \notag \\ &= \frac{2 \sin \omega \sin \beta_0}{\pi} \int_0^{\arctan \frac{R}{h}} \sin^2 t dt = \frac{\sin \omega \sin \beta_0}{\pi} \int_0^{\arctan \frac{R}{h}} 1 - \cos 2t dt \notag \\ &= \frac{\sin \omega \sin \beta_0}{\pi} \left[t - \frac{1}{2} \sin 2t \right]^{\arctan \frac{R}{h}}_0 = \frac{\sin \omega \sin \beta_0}{\pi} \left[t - \frac{\tan t}{1+ \tan^2 t} \right]^{\arctan \frac{R}{h}}_0 \notag \\ &= \frac{\sin \omega \sin \beta_0}{\pi} \left( \arctan \frac{R}{h} - \frac{Rh}{R^2 + h^2} \right) \end{align}$$

Second term:

$$\begin{align} &\int^R_0 \frac{2\beta_0 rh^2 \cos \omega}{\pi (r^2 + h^2)^2}dr = \left[ - \frac{\beta_0 h^2 \cos \omega}{\pi (r^2 + h^2)} \right]^R_0 = \frac{\beta_0 \cos \omega}{\pi} \frac{R^2}{R^2 + h^2} \notag \\ &= \frac{\beta_0 \cos \omega}{\pi} \frac{1}{1 + (h/R)^2} \end{align}$$

Third term:

$$\begin{align} &\int_0^{R \sin \beta_0} \int^{\frac{y}{\tan \beta_0}}_{R \cos \beta_0} \frac{2xh \sin \omega}{\pi (x^2 + y^2 + h^2)^2} dx dy = \int_0^{R \sin \beta_0} \left[ - \frac{h \sin \omega}{\pi (x^2 + y^2 + h^2)} \right]^{\frac{y}{\tan \beta_0}}_{R \cos \beta_0} dy \notag \\ &= \frac{h \sin \omega}{\pi}\int_0^{R \sin \beta_0} \left( \frac{1}{y^2 + h^2 + R^2 \cos^2 \beta_0} - \frac{1}{\frac{y^2}{\sin^2 \beta_0} + h^2} \right) dy \notag \\ &= \frac{h \sin \omega}{\pi}\int_0^{R \sin \beta_0} \frac{1}{\frac{y^2}{h^2 + R^2 \cos^2 \beta_0} + 1} \frac{dy}{h^2 + R^2 \cos^2 \beta_0} - \frac{h \sin \omega}{\pi} \int_0^{R \sin \beta_0} \frac{1}{\frac{y^2}{h^2 \sin^2 \beta_0} + 1} \frac{dy}{h^2} \notag \\ &= \frac{h \sin \omega}{\pi} \int_0^{\arctan \frac{R\sin \beta_0}{\sqrt{h^2 + R^2 \cos^2 \beta_0}}} \frac{du}{\sqrt{h^2 + R^2 \cos^2 \beta_0}} - \frac{h \sin \omega}{\pi} \int_0^{\arctan \frac{R}{h}} \frac{\sin \beta_0}{h} dv \notag \\ &= \frac{h \sin \omega}{\pi \sqrt{h^2 + R^2 \cos^2 \beta_0}} \arctan \frac{R\sin \beta_0}{\sqrt{h^2 + R^2 \cos^2 \beta_0}} - \frac{\sin \omega \sin \beta_0}{\pi} \arctan \frac{R}{h} \end{align}$$

Fourth term:

First, we perform integration in the $Y$ direction.

$$\begin{align} &\int_0^{x \tan \beta_0} \frac{1}{(y^2 + x^2 + h^2)^2} dy \notag \\ &= \int_0^{x \tan \beta_0} \frac{\frac{1}{(x^2 + h^2)^2}}{(\frac{y^2}{x^2 + h^2} + 1)^2} dy = \frac{\sqrt{x^2 + h^2}}{(x^2 + h^2)^2} \int_0^{\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}}} \frac{1}{(u^2 + 1)^2} du \notag \\ &= \frac{\sqrt{x^2 + h^2}}{(x^2 + h^2)^2} \int_0^{\arctan(\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}})} \frac{1}{(\tan^2 t + 1)^2} \frac{1}{\cos^2 t}dt \notag \\ &= \frac{\sqrt{x^2 + h^2}}{(x^2 + h^2)^2} \int_0^{\arctan(\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}})} \cos^2 t dt = \frac{\sqrt{x^2 + h^2}}{2(x^2 + h^2)^2} \int_0^{\arctan(\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}})} \cos 2t + 1 dt \notag \\ &= \frac{\sqrt{x^2 + h^2}}{2(x^2 + h^2)^2} \left[ \frac{1}{2}\sin 2t + t\right]_0^{\arctan(\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}})} = \frac{\sqrt{x^2 + h^2}}{2(x^2 + h^2)^2} \left[ \frac{\tan t}{\tan^2 t + 1} + t\right]_0^{\arctan(\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}})} \notag \\ &= \frac{\sqrt{x^2 + h^2}}{2(x^2 + h^2)^2} \left\{ \frac{\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}}}{\frac{x^2 \tan^2 \beta_0}{x^2 + h^2} + 1} + \arctan \left(\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}}\right)\right\} \notag \\ &= \frac{1}{2} (x^2 + h^2)^{-1} \frac{x \tan \beta_0}{x^2 (\tan^2 \beta_0 + 1) + h^2} + \frac{1}{2} (x^2 + h^2)^{-\frac{3}{2}} \arctan \frac{x \tan \beta_0}{\sqrt{x^2 + h^2}} \end{align}$$

The parameter transformation has been performed as shown in Eqs. (11) and (12).

$$\begin{align} &\frac{x \tan \beta_0}{\sqrt{x^2 + h^2}} = \tan u \\ &h^2 \tan \beta_0 (x^2 + h^2)^{-\frac{3}{2}} dx = \frac{du}{\cos^2 u} \end{align}$$

Then, we integrate two terms in Eq. (10) in the $X$ direction.

$$\begin{align} &\int^0_{R \cos \beta_0} \frac{1}{2} (x^2 + h^2)^{-\frac{3}{2}} \arctan \frac{x \tan \beta_0}{\sqrt{x^2 + h^2}} dx \notag\\ &= \int^0_{\arctan \left( \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \right)} \frac{u du}{2 h^2 \tan \beta_0 \cos^2 u} \notag \\ &= \frac{1}{2h^2 \tan \beta_0}\left[ u \tan u + \log (\cos u) \right]^0_{\arctan \left( \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \right)} \notag \\ &= -\frac{1}{2h^2 \tan \beta_0} \left[ \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \arctan \left( \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \right)\right. \notag \\ &\hspace{11pt}\times \left.\log \left\{ \cos \left( \arctan \left( \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \right) \right) \right\} \right] \end{align}$$ $$\begin{align} &\int^0_{R \cos \beta_0} \frac{1}{2} (x^2 + h^2)^{-1} \frac{x \tan \beta_0}{x^2 (\tan^2 \beta_0 + 1) + h^2} dx \notag \\ &= \int^0_{R \cos \beta_0} \frac{1}{4 h^2 \tan \beta_0} \left\{ \frac{2x(\tan^2 \beta_0 + 1)}{(\tan^2 \beta_0 + 1) x^2 + h^2} - \frac{2x}{x^2 + h^2} \right\} dx \notag \\ &= \frac{1}{4 h^2 \tan \beta_0} \left[ \log \left( 1 + \frac{x^2 \tan^2 \beta_0}{x^2 + h^2} \right) \right]^0_{R \cos \beta_0} \notag \\ &= - \frac{1}{4 h^2 \tan \beta_0} \log \left( 1 + \frac{R^2 \sin^2 \beta_0}{R^2 \cos^2 \beta_0 + h^2} \right) \end{align}$$

Finally, the fourth term can be expressed as follows:

$$\begin{align} &\int^0_{R \cos \beta_0} \int_0^{x \tan \beta_0} \frac{2h^2 \cos \omega}{\pi (x^2 + y^2 + h^2)^2} dy dx \notag \\ &= -\frac{\cos \omega}{\pi \tan \beta_0} \left[ \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \arctan \left( \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \right)\right. \notag \\ &\hspace{70pt}\times \left.\log \left\{ \cos \left( \arctan \left( \frac{R \sin \beta_0}{\sqrt{R^2 \cos^2 \beta_0 + h^2}} \right) \right) \right\} \right] \notag \\ &\hspace{11pt}- \frac{\cos \omega}{2 \pi \tan \beta_0} \log \left( 1 + \frac{R^2 \sin^2 \beta_0}{R^2 \cos^2 \beta_0 + h^2} \right) \end{align}$$

In this case, we managed to complete the integration, thanks to the shape simplicity of the disk. However, performing the double integration is highly complex, and it would be very difficult to obtain a closed-form solution for more complicated geometries.

View Factor Evaluation based on the Line Integration

In some cases, the area integral can be transformed into a line integral using Stokes’ theorem. This is not always possible, but it can be applied to the calculation of the view factor. Stokes’ theorem is expressed as shown in Eq. (16) as vector form, and Eq. (17) as explicit parametric form.

$$\begin{equation} \iint_{\Omega} \nabla \times \boldsymbol{F} \cdot d\boldsymbol{s} = \int_{\partial \Omega} \boldsymbol{F} \cdot d \boldsymbol{l} \end{equation}$$ $$\begin{align} \iint_{\Omega} \left[ l \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} \right) + m \left( \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} \right) + n \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \right] ds \notag \\ = \int_{\partial \Omega} \left( P \frac{dx}{dt} + Q \frac{dy}{dt} + R \frac{dz}{dt} \right) dt \end{align}$$

The view factor in question is expressed as shown in Eq. (18).

$$\begin{align} F &= \int_A \frac{\cos \psi \cos \lambda}{\pi S^2} dA = \int \frac{(\boldsymbol{x} \cdot \boldsymbol{n}_1) \times (-\boldsymbol{x} \cdot \boldsymbol{n}_2)}{\pi S^4} dA \notag \\ &= \int (-l_2 x f - m_2 y f - n_2 z f)\ dA, \quad \mathrm{where}\quad f = \frac{1}{\pi S^4} (l_1 x + m_1 y + n_1 z) \end{align}$$

Comparing the expressions in Eqs. (17) and (18), the following relationships must hold for the application of Stokes’ theorem.

$$\begin{align} \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = - x f, \quad \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = - y f, \quad \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = - z f \end{align}$$

While this choice of $P$, $Q$, and $R$ might appear arbitrary, it conveniently fulfills the conditions.

$$\begin{align*} P = \frac{-m_1 z + n_1 y}{2 \pi S^2}, \hspace{10pt} Q = \frac{-n_1 x + l_1 z}{2 \pi S^2}, \hspace{10pt} R = \frac{-l_1 y + m_1 x}{2 \pi S^2} \end{align*}$$

As an example, we can verify the first condition in Eq. (19) as shown in Eqs. (20)—(22).

$$\begin{align} \frac{\partial R}{\partial y} &= \frac{\partial}{\partial y} \left(\frac{-l_1 y + m_1 x}{2 \pi S^2} \right) = \frac{-l_1}{2 \pi S^2} - \frac{(-l_1y + m_1x)y}{\pi S^4} \\ \frac{\partial Q}{\partial z} &= \frac{\partial}{\partial z} \left(\frac{-n_1 x + l_1 z}{2 \pi S^2} \right) = \frac{l_1}{2 \pi S^2} - \frac{(-n_1x + l_1z)z}{\pi S^4} \end{align}$$ $$\begin{align} \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} &= \frac{-l_1(x^2 + y^2 + z^2)}{\pi S^4} + \frac{l_1y^2 - m_1xy - n_1xz + l_1z^2}{\pi S^4} \notag \\ &= \frac{-x (l_1 x + m_1 y + n_1 z)}{\pi S^4} = -xf \end{align}$$

Since we have specified appropriate $P$, $Q$, and $R$, we can express the view factor as a line integral.

$$\begin{align} F &= \frac{1}{2 \pi} \int \left( \frac{-m_1 z + n_1 y}{S^2} dx + \frac{-n_1 x + l_1 z}{S^2} dy + \frac{-l_1 y + m_1 x}{S^2} dz \right) \notag \\ &= l_1 \int \frac{z dy - y dz}{2 \pi S^2} + m_1 \int \frac{x dz - z dx}{2 \pi S^2} + n_1 \int \frac{y dx - x dy}{2 \pi S^2} \end{align}$$ $$\begin{align*} &\left[ \begin{array}{c}l_1 \\ m_1 \\ n_1\end{array} \right] =\left[ \begin{array}{c}\sin \omega \\ 0 \\ \cos \omega\end{array} \right] &&:\quad \text{Infinitesimal surface direction} \\ &\left[ \begin{array}{c}x \\ y \\ z\end{array} \right] =\left[ \begin{array}{c}R \cos \alpha \\ R \sin \alpha \\ h\end{array} \right] &&:\quad \text{Line on the disk edge} \\ &\left[ \begin{array}{c}x \\ y \\ z\end{array} \right] =\left[ \begin{array}{c}R \cos \alpha_0 \\ y \\ h\end{array} \right] &&:\quad \text{Line of the view edge} \end{align*}$$

Figure 3: Disk View Factor Calculation by Line Integration.

Now, we are ready to perform the view factor calculation by the line integral. The vector $(l_2, m_2, n_2)$ representing the orientation of the disk is directed downward, and we will integrate in a clockwise direction with respect to this vector.

$$\begin{align} &F = l_1 \int \frac{z dy - y dz}{2 \pi S^2} + n_1 \int \frac{y dx - x dy}{2 \pi S^2} \notag \\ &= \sin \omega \left( \int_{\alpha_0}^{-\alpha_0} \frac{h (R \cos \alpha)}{2 \pi (R^2 + h^2)} d\alpha + \int_{-R \sin \alpha_0}^{R \sin \alpha_0} \frac{h}{2 \pi (R^2 \cos^2 \alpha_0 + y^2 + h^2)} dy \right) \notag \\ &\hspace{11pt}+ \cos \omega \left( \int_{\alpha_0}^{-\alpha_0} \frac{R \sin \alpha (-R \sin \alpha) - R \cos \alpha (R \cos \alpha)}{2 \pi (R^2 + h^2)} d\alpha + \int_{-R \sin \alpha_0}^{R \sin \alpha_0} \frac{- R \cos \alpha_0}{2 \pi (R^2 \cos^2 \alpha_0 + y^2 + h^2)} dy \right) \notag \\ &= \frac{Rh \sin \omega}{2 \pi (R^2 + h^2)} \int^{-\alpha_0}_{\alpha_0} \cos \alpha d\alpha + \frac{h \sin \omega}{2 \pi} \int_{\arctan(-\frac{R \sin \alpha_0}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}})}^{\arctan(\frac{R \sin \alpha_0}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}})} \frac{dt}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}} \notag \\ &\hspace{11pt}- \frac{R^2}{2 \pi (R^2 + h^2)} \int_{\alpha_0}^{-\alpha_0} d\alpha - \frac{R \cos \omega \cos \alpha_0}{2 \pi} \int_{\arctan(-\frac{R \sin \alpha_0}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}})}^{\arctan(\frac{R \sin \alpha_0}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}})} \frac{dt}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}} \notag \\ &= - \frac{Rh \sin \omega \sin \alpha_0}{\pi (R^2 + h^2)} + \frac{h \sin \omega}{\pi \sqrt{R^2 \cos^2 \alpha_0 + h^2}} \arctan \left( \frac{R \sin \alpha_0}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}} \right) \notag \\ &\hspace{11pt}+ \frac{R^2 \alpha_0 \cos\omega}{\pi (R^2 + h^2)} - \frac{R \cos \omega \cos \alpha_0}{\pi \sqrt{R^2 \cos^2 \alpha_0 + h^2}} \arctan \left( \frac{R \sin \alpha_0}{\sqrt{R^2 \cos^2 \alpha_0 + h^2}} \right) \end{align}$$

With this approach, we have successfully derived the view factor expression for a disk from a plate element with reduced complexity.

Disk View Factor from a Plate Element with Offset

As another case of disk view factor from a plate element, let’s consider the scenario where the plate element is offset from the central axis of the disk. As shown in Figure 4, the orientation of the plate element is parallel to the disk.

Figure 4: Plate Element to Disk in Parallel Plane. Element is Offset from the Disk Center.

Assuming the origin is at the plate element, the edge of the disk can be expressed as follows.

$$\begin{align} \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} R \cos\alpha - a \\ R \sin\alpha \\ h \end{array} \right], \quad \frac{d}{d\alpha} \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] = \left[ \begin{array}{c} -R \sin\alpha \\ R \cos\alpha \\ 0 \end{array} \right] \end{align}$$

The distance $S$ from a point on the edge of the disk to the infinitesimal surface is expressed as follows.

$$\begin{align} S^2 &= (R \cos \alpha - a)^2 + (R \sin \alpha)^2 + h^2 \notag \\ &= R^2 + a^2 + h^2 -2aR \cos\alpha \end{align}$$

As the orientation of the plate element is set to $(0,0,1)$, the view factor can be calculated as follows.

$$\begin{align} &F_\mathrm{offset} = \int \frac{ydx - xdy}{2\pi S^2} = \int_{\pi}^{-\pi} \frac{-R^2 + aR \cos\alpha}{2\pi (R^2 + a^2 + h^2 -2aR \cos\alpha)}d\alpha \notag \\ &= \frac{R^2}{2\pi} \int_{-\pi}^{\pi} \frac{d\alpha}{R^2 + a^2 + h^2 - 2aR \cos\alpha} - \frac{aR}{2\pi} \int_{-\pi}^{\pi} \frac{\cos\alpha ~d\alpha}{R^2 + a^2 + h^2 - 2aR \cos\alpha} \end{align}$$

The first term in (27) can be calculated as follows:

$$\begin{align} &\frac{R^2}{2\pi} \int_{-\pi}^{\pi} \frac{d\alpha}{R^2 + a^2 + h^2 - 2aR \cos\alpha} \notag \\ &=\frac{R^2}{2\pi} \int_{-\infty}^{\infty} \frac{1}{R^2 + a^2 + h^2 - 2aR \frac{1-t^2}{1+t^2}} \frac{2}{1+t^2}dt \notag \\ &= \frac{R^2}{\pi} \int_{-\infty}^{\infty} \frac{1}{(R^2 + a^2 + h^2 - 2aR) + (R^2 + a^2 + h^2 + 2aR)t^2}dt \notag \\ &= \frac{R^2}{\pi(R^2 + a^2 + h^2 + 2aR)} \int_{-\infty}^{\infty} \frac{1}{A^2 + t^2}dt \notag \\ &= \frac{R^2 A}{\pi(R^2 + a^2 + h^2 - 2aR)} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \notag = \frac{R^2A}{R^2 + a^2 + h^2 - 2aR} \notag \\ &= \frac{R^2}{\sqrt{(R^2 + a^2 + h^2)^2 - 4a^2R^2}} \end{align}$$

The parameter $A$ and the substitution used in the integration are presented as shown below.

$$\begin{align} t = A \tan\theta, \quad dt = \frac{A~d\theta}{\cos^2\theta}, \quad A = \sqrt{\frac{R^2 + a^2 + h^2 - 2aR}{R^2 + a^2 + h^2 + 2aR}} \end{align}$$

The second term in (27) can be calculated as follows:

$$\begin{align} &\frac{aR}{2\pi} \int_{-\pi}^{\pi} \frac{\cos\alpha ~d\alpha}{R^2 + a^2 + h^2 - 2aR \cos\alpha} \notag \\ &=\frac{aR}{2\pi} \int_{-\infty}^{\infty} \frac{\frac{1-t^2}{1+t^2}}{R^2 + a^2 + h^2 - 2aR \frac{1-t^2}{1+t^2}} \frac{2}{1+t^2}dt \notag \\ &=\frac{aR}{\pi} \int_{-\infty}^{\infty} \frac{1-t^2}{(R^2 + a^2 + h^2 - 2aR) + (R^2 + a^2 + h^2 + 2aR)t^2} \frac{1}{1+t^2}dt \notag \\ &=\frac{aR}{\pi(R^2 + a^2 + h^2 + 2aR)} \int_{-\infty}^{\infty} \frac{1-t^2}{(A^2 + t^2)(1+t^2)}dt \notag \\ &=\frac{aR}{\pi(R^2 + a^2 + h^2 + 2aR)} \int_{-\infty}^{\infty} \left(\frac{1+A^2}{1-A^2} \frac{1}{t^2 + A^2} - \frac{2}{1-A^2} \frac{1}{t^2 + 1}\right) dt \notag \\ &=\frac{aRA}{R^2 + a^2 + h^2 - 2aR} \frac{1+A^2}{1-A^2} - \frac{aR}{R^2 + a^2 + h^2 + 2aR} \frac{2}{1-A^2} \notag \\ &=\frac{1}{2}\frac{R^2 + a^2 + h^2}{\sqrt{(R^2 + a^2 + h^2)^2 - 4a^2R^2}} - \frac{1}{2} \end{align}$$

Finally, the disk view factor from a plate element with an offset is expresseed by the following equation.

$$\begin{align} F_\mathrm{offset} &= \frac{1}{2} - \frac{1}{2}\frac{a^2 + h^2 - R^2}{\sqrt{(R^2 + a^2 + h^2)^2 - 4a^2R^2}} \end{align}$$

References

1. John R. Howell, M. Pinar Mengüç, “Radiative transfer configuration factor catalog: A listing of relations for common geometries”, Journal of Quantitative Spectroscopy and Radiative Transfer, Volume 112, Issue 5, 2011, Pages 910-912, doi: 10.1016/j.jqsrt.2010.10.002
2. A Catalog of Configuration Factors, 3rd Edition, https://www.thermalradiation.net/indexCat.html
3. View Factor Calculator, https://sterad.net